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Number of resistors for potential divider?

  1. Jul 20, 2011 #1
    Not sure whether to post in mathematics or physics section.

    A circuit contains a cell of unit E.M.F. Use any combination of identical resistors in series and/or parrallel in order to create a voltage of x across a voltmeter connected anywhere on the circuit. Assume that there is no internal resistance or resistance in the wires, and the voltmeter does not draw any current. What is the easiest way to find the minimum number of resistors required for a given value of x (x<1 and x is rational, otherwise there is no solution)?
    Last edited: Jul 21, 2011
  2. jcsd
  3. Jul 20, 2011 #2
    You lost me
  4. Jul 20, 2011 #3
    I don't think there is an easy way to find the minimum numbers. You can make any fraction p/q, with q resistors in series, where you connect the voltmeter across p of them, but this is not always the smallest number.

    with four resistances you can make a voltage of 1/5 by putting 2 in series to get a 2 ohm resistance, and 2 in parallel to get 1/2 ohm. If you make a voltage divider from 2 ohm and 1/2 ohm you get 1/2 / (2 + 1/2) = 1/5.
  5. Jul 21, 2011 #4
    The only pattern I could be sure of is that the numbers of resistors for x=a/b is the same as for x=(b-a)/b. It also looks like b resistors are required when x=1/b and b is prime.
    A few values I have found:
    voltage no. of resistors
    1/2 2
    1/3 3
    1/4 4
    1/5 5
    2/5 4
    1/6 4
    1/7 7
    2/7 5
    3/7 5
    1/8 8
    3/8 7
    1/10 5
    2/13 8
    7/13 7
    2/19 19
    19/23 13
    2/31 17
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